python - Checking a nested dictionary using a dot notation string "a.b.c.d.e", automatically create missing levels

14
votes

Given the following dictionary:

d = {"a":{"b":{"c":"winning!"}}}

I have this string (from an external source, and I can't change this metaphor).

k = "a.b.c"

I need to determine if the dictionary has the key 'c', so I can add it if it doesn't.

This works swimmingly for retrieving a dot notation value:

reduce(dict.get, key.split("."), d)

but I can't figure out how to 'reduce' a has_key check or anything like that.

My ultimate problem is this: given "a.b.c.d.e", I need to create all the elements necessary in the dictionary, but not stomp them if they already exist.

4
pythondictionarynesteddefaultdict
Yes, just accept your own answers to your questions. That's perfectly OK (even desired, so people who have the same question will see that there exists an answer). - Tim Pietzcker
By any chance is the source actually Cocoa KVC? And, if so, can you just use PyObjC instead of pure Python? (Besides being simpler, that would also guarantee that the keys mean exactly the same thing they're intended to, even if there are bugs in other people's code…) - abarnert
What is so special in using dot notation 'as a string' to search through dictionary? One has to use some additional function for it. What about to search multi-dimentional dicts JS style, without any strings x = conf.a.b.c. I don't know if it is possible in Python. But using a string for it is not cool - Green
That 'reduce' fragment helped me. Thanks! - Alain Collins
Near-duplicate, although from 2018 and presumably you asked about Python 2.x not 3.x Set Python dict items recursively, when given a compound key 'foo.bar.baz' - smci

4 Answers

13
votes

... or using recursion:

def put(d, keys, item):
    if "." in keys:
        key, rest = keys.split(".", 1)
        if key not in d:
            d[key] = {}
        put(d[key], rest, item)
    else:
        d[keys] = item

def get(d, keys):
    if "." in keys:
        key, rest = keys.split(".", 1)
        return get(d[key], rest)
    else:
        return d[keys]
26
votes

You could use an infinite, nested defaultdict:

>>> from collections import defaultdict
>>> infinitedict = lambda: defaultdict(infinitedict)
>>> d = infinitedict()
>>> d['key1']['key2']['key3']['key4']['key5'] = 'test'
>>> d['key1']['key2']['key3']['key4']['key5']
'test'

Given your dotted string, here's what you can do:

>>> import operator
>>> keys = "a.b.c".split(".")
>>> lastplace = reduce(operator.getitem, keys[:-1], d)
>>> lastplace.has_key(keys[-1])
False

You can set a value:

>>> lastplace[keys[-1]] = "something"
>>> reduce(operator.getitem, keys, d)
'something'
>>> d['a']['b']['c']
'something'
3
votes

How about an iterative approach?

def create_keys(d, keys):
    for k in keys.split("."):
        if not k in d: d[k] = {}  #if the key isn't there yet add it to d
        d = d[k]                  #go one level down and repeat

If you need the last key value to map to anything else than a dictionary you could pass the value as an additional argument and set this after the loop:

def create_keys(d, keys, value):
    keys = keys.split(".")
    for k in keys[:-1]:
        if not k in d: d[k] = {}
        d = d[k]            
    d[keys[-1]] = value
0
votes
d = {"a":{}}
k = "a.b.c".split(".")

def f(d, i):
    if i >= len(k):
        return "winning!"
    c = k[i]
    d[c] = f(d.get(c, {}), i + 1)
    return d

print f(d, 0)
"{'a': {'b': {'c': 'winning!'}}}"